[릿코드 / js] 다시풀기 1631. Path With Minimum Effort

var minimumEffortPath = function(H) {
    let y = H.length, x = H[0].length, last = x*y-1,
        vis = new Uint8Array(last+1),
        dist = new Uint32Array(last+1).fill(1000001), 
        heap = [,], node = 0, path = 0, i, j, cell
    const heapify = (node, k, l) => {
        let i = heap.length, par = i >> 1
        dist[node] = Math.min(dist[node], Math.max(path, Math.abs(cell - H[k][l])))
        heap.push([node,dist[node]])
        while (par && heap[par][1] > heap[i][1]) {
            [heap[par], heap[i]] = [heap[i], heap[par]]
            i = par, par = i >> 1
        }
    }
    const extract = () => {
        let min = heap[1], left, right,
            i = 1, child = heap[3] && heap[3][1] < heap[2][1] ? 3 : 2
        heap[1] = heap.pop()
        while (heap[child] && heap[i][1] > heap[child][1]) {
            [heap[i], heap[child]] = [heap[child], heap[i]]
            i = child, left = i << 1, right = left + 1
            child = heap[right] && heap[right][1] < heap[left][1] ? right : left
        }
        return min
    }
     while (node !== last) {
        i = ~~(node / x), j = node % x, cell = H[i][j]
        if (i > 0 && !vis[node-x]) heapify(node-x, i-1, j)
        if (i < y-1 && !vis[node+x]) heapify(node+x, i+1, j)
        if (j > 0 && !vis[node-1]) heapify(node-1, i, j-1)
        if (j < x-1 && !vis[node+1]) heapify(node+1, i, j+1)
        vis[node] = 1
        while (vis[node]) [node,path] = extract()
    }
    return path
};